schroedinger equation - Explicit nontrivial examples in quantum mechanics: ground state degeneracy

Ground state degeneracy occurs whenever there exists a unitary operator which acts non-trivially on a ground state and commutes with the Hamiltonian of the system.

I just want to find a potential $V(\mathbf{r})$, not necessary the central potential, such that Schrodinger equation in d-dimensional (no internal degrees of freedom like spin) $$E\Psi (\mathbf {r})=\left[{\frac {-\hbar ^{2}}{2\mu }}\nabla ^{2}+V(\mathbf {r} )\right]\Psi (\mathbf {r} )$$ has degenerate ground state.

I've tried many ways, but failed.

1 For example, given a potential $V(\mathbf{r})$, and solve eigenenergy $E_0, E_1\cdots$. I want to construct $H'=H-(E_1-E_0)|1\rangle\langle1|$, but this part $(E_1-E_0)|1\rangle\langle1|$ in position representation is not a local potential.

2 Certainly, it's easy to construct the finite dimensional quantum system have ground state degeneracy, that is we can write Hamiltonian matrix as a diagonal matrix with multiple lowest eigenvalues $H=diag(E_0,E_0,E_1 )$. But I don't want this trivial way.

3 It's also easy to construct quantum mechanical system with internal degree of freedom like spin. And internal degrees of freedom don't have dynamics. For example, hydrogen model with spin degree of freedom. For lowest energy $n=1,l=0$, we can have $s=\pm1/2$ with same energy. This way is also trivial.

4 And we know the scattering state in 1-dim has continuous spectra and every state is double degenerate. I want to construct $V(\mathbf{r})\ge0$ such that $$\lim_{|\mathbf{r}|\rightarrow\infty}V(\mathbf{r})\rightarrow 0$$ Even though all $E>0$ have degeneracy, $E=0$ is still unique.

5 Certainly, in 1-dim if $V(x)$ is a double infinite deep potential, we can have degenerate ground state. But this example is also trivial.

6 The potential with spontaneous broken symmetries, e.g. $V(x)= -x^2 +x^4$, is also impossible. There is a enegry gap between even parity and odd parity.

So my question is, apart from above trivial examples whether we can construct a example, that is in d-dim a particle without internal d.o.f, like spin, can have degenerate ground state in some potential.

This question may be a question in partial differential function. If such $V(\mathbf{r})$ does not exist, how to prove.

Answer

For Hamitonian operator like this form $-\Delta +V(x)$, the ground state is always non-degeneracy in $n$-dim if the potential is continuous and bounded from below and let $-\Delta +V(x)$ be essentially self-adjoint. You can see the proof in Page 51 James Glimm and Arthur Jaffe's Quantum Physics. Or see the proof.

If you don't limit the Hamitonian to this form( $-\Delta +V(x)$), then if you put magnetic field then it's easy to construct the degeneracy ground state. see Landau level.